REG Q01=ADV Q01 Semester 1, Y2011 - Home - The …sydney.edu.au/science/physics/pdfs/current/jphys/exa… ·  · 2011-09-14A projectile of mass 2m is fired with a speed u at an angle

REG_Q01=ADV_Q01

Semester 1, Y2011 Question 1

A large wooden turntable in the shape of a flat uniform disk is freely rotating about a vertical axis through its centre. A parachutist descends vertically and makes a soft landing on the turntable, without slipping, at a point near its outer rim.

(a) Will the angular speed of the turntable increase or decrease after the parachutist lands? Briefly justify your answer.

(b) Will the kinetic energy of the system (turntable + parachutist) increase or decrease in the process? Briefly explain your answer.

(5 marks)

Solution

(a) Angular momentum L Iω= will be conserved in the process. The addition of the parachutist will increase I and hence ω will decrease. (2 marks) (b) We have 1 1 2 2I Iω ω= where, 1I is the initial moment of inertia of the turntable and 1ω is its angular velocity, 2I is the moment of inertia of the turntable plus parachutist and 2ω is its angular velocity. Therefore

12 1 1

2

II

ω ω ω= < since 2 1.I I>

The kinetic energy is

22 2 21 1

2 2 2 2 1 1 122 2

11

2

1 1 12 2 2

I IK I I II I

I KI

ω ω ω= = =

=.

So 2 1K K< since 2 1.I I> Hence kinetic energy decreases. Friction will do net negative work on the system (turntable + parachutist) in the process of accelerating the parachutist in the horizontal direction. The friction would in reality most likely arise from slipping between the parachutist and the turntable. But it could also arise from ‘deformation’ of the parachutist as he/she is accelerated to the velocity of the turntable. Energy is lost as a result. (3 marks)

ADV_Q02

Semester 1, Y2011 Question 2

A projectile of mass 2 m is fired with a speed u at an angle θ above the horizontal. At the highest point of its trajectory the projectile explodes into two fragments of equal mass, one of which falls vertically with zero initial speed. The range of the projectile if it had not exploded would have been R . Ignore air resistance.

(a) Is momentum conserved during the explosion? Briefly explain your answer.

(b) How far from the point of firing does the other fragment land, assuming that the terrain is level? Express your answer in terms of .R

(c) By what factor has the kinetic energy of the projectile changed just after the explosion when compared with its value just before the explosion?

(5 marks)

Solution (a) Momentum is conserved during the explosion because the forces acting on the two fragments as a result of the explosion are internal to the projectile system. (1 mark) (b) The backward fragment falls at distance 1 / 2x R= from point of firing since at the highest point the projectile has travelled half the range. The centre of mass falls at distance R since the centre of mass is unaffected by the explosion, which is internal to the projectile system. Taking the origin at the launch point and from the centre of mass formula we derive the value for

2x , the point at which the forward fragment lands.

1 1 2 2

1 21

22

122

32 2

22

cmm x m xx

m mm R mxR

mR R x

x R

+=

++

=

= +

=

The second fragment lands at 3 / 2R from the launch point. Alternate method using conservation of momentum Take xu as the horizontal velocity of the projectile just before the explosion and xv as the velocity of the forward fragment just after the explosion. Using conservation of momentum we have 2 (0)

2x x

x x

mu m mvv u

= +⇒ =

We can treat the horizontal and vertical motions separately. The project would have travelled a further distance / 2R from the peak of the path until it hit the ground. The forward fragment travels at twice the horizontal speed and hence goes a distance 2 / 2R R= until it hits the ground. The total distance that it has travelled from the origin is therefore / 2 3 / 2R R R+ = (2 marks) (c) Kinetic energy before explosion: 2 21

2 2 x xmu mu= Kinetic energy after explosion: 21

2 0xm v + From conservation of momentum at explosion:

2 02x x

x x

mu mvv u

= +=

Therefore kinetic energy after explosion: ( )22 21 12 2 2 2x x xmv m u mu= =

Therefore kinetic energy has increased by a factor 2. (2marks)

ADV_Q03

Semester 1, Y2011 Question 3

A yo-yo of mass m has an axle of radius b and a spool of radius R . Its moment of inertia about

an axis passing through the centre of the yo-yo is 212

I m R= . The coefficient of static friction

between the yo-yo and the table is sμ . The string, wrapped around the spool as shown in the figure, is pulled to the right with a horizontal force F . Depending on the force applied the yo-yo will either not move, roll to the right, or slip to the right.

(a) Show that the net force netF when the yo-yo is at the point of slipping is given by

.net sF F m gμ= −

(b) Show that the net clockwise torque netτ around the centre of rotation when the yo-yo is at the point of slipping is given by

.net s m g R F bτ μ= −

(c) Show the maximum magnitude of F for which the yo-yo will roll without slipping is given by

32s

RF m gR b

μ⎛ ⎞

= ⎜ ⎟+⎝ ⎠.

Hint: Consider the linear and angular accelerations at the point of contact of the yo-yo and the table.

(5 marks)

Solution

(a) Taking upwards as the positive direction, the gravity force acting on the yo-yo is:

gF m g= − .

The yo-yo is in equilibrium in the vertical direction and so the normal reaction force from the table acting (upwards) on the yo-yo is: .N m g=

At the point of almost slipping, we have a frictional force f given by:

s sf N m gμ μ= =

acting to the left in the diagram as we are told that the yo-yo will either roll or slip to the right.

The net force acting on the yo-yo is therefore

.net sF F f F m gμ= − = − (1)

(1 mark) (b) Taking the clockwise direction as positive and considering torques acting around the centre of rotation of the yo-yo we have: Torque due to pulling force pull F bτ = − . Torque due to frictional force fric f Rτ = At the point of almost slipping, we have:

s sf N m gμ μ= =

Therefore the net torque in the clockwise direction is:

.net sf R F b m g F bτ μ= − = − (2)

(1 mark)

(c) From equation (1) the linear acceleration is given by:

( )net F fFam m

−= =

(½ mark) From equation (2) the angular acceleration is given by:

( )2 2

20.5

f R F bf R F bI m R m Rτα

−−= = =

(1 mark) For the contact point with the floor the linear and the angular acceleration are related by a Rα= , (½ mark) and we have

( )2 f R F bF fm m R

−−=

2 22 2

32

32s

F R f R f R F bF R F b f R f R

f RFR b

RF m gR b

μ

− = − ⇒+ = + ⇒

= ⇒+

⎛ ⎞= ⎜ ⎟+⎝ ⎠

where we have set sf m gμ= for the yo-yo on the verge of slipping. (1 mark)

ADV_Q04

Semester 1, Y2011 Question 4 (a) In some household air conditioners used in dry climates, air is cooled by blowing it

through a water-soaked filter. Briefly explain how this cools the air. Explain why these cooling systems do not work as well in a high-humidity climate.

(b) How much faster are molecules moving, on average, in air at 100 CT = than in air at 25 CT = ?

(c) A CO2 fire extinguisher contains pressurised carbon dioxide gas that emerges as solid dry ice when released. Briefly explain why it emerges in this state.

(d) A growing plant creates a highly complex and organised structure out of simple things such as sunlight, air, water and trace minerals. Does this violate the second law of thermodynamics? Explain your reasoning.

(5 marks)

Solution (a) Evaporative cooling – heat is transferred from the hot air to the water, which evaporates, removing latent heat of vaporisation (1 mark). In a high-humidity climate, evaporation is less efficient because the air is saturated with water vapour, so an evaporative cooling system would not work as well as in a dry climate (½ mark). (b)

21 32 2tr avK m v k T= = is the average (translational) kinetic energy of the gas molecules, so

avv T∼ (1 mark).

Therefore the ratio of speeds is 373 1.12298

KK

= , or 12% faster (½ mark).

(c) Adiabatic expansion (½ mark) – the gas is released so quickly that there is no heat transfer with the surroundings, so the gas cools to below the CO2 freezing temperature (½ mark). (note that at a pressure of 1 atm, CO2 sublimates) (d) No it does not violate the 2nd law because this is an irreversible process (½ mark) and hence, total entropy must increase. In consuming the raw materials needed for growth, this process has resulted in less energy being available in the universe to convert to mechanical work, so the entropy of the universe has increased more than the entropy of the plant has decreased (½ mark).

REG_Q05=ADV_Q05

Semester 1, Y2011 Question 5

The diagram above shows a wire with its left end attached to an oscillating pin that is fixed to a table. The wire extends to the right over a frictionless pulley and is attached to a hanging weight of mass 5.00 kg . The distance between the pin and the pulley is 1.00 m and the linear density of the wire is 10.100kg.m− . The wire is made to vibrate when the pin is connected to a frequency generator causing it to oscillate up and down. The oscillating pin generates a sinusoidal travelling transverse wave in the wire with a frequency of 40.0 Hz and amplitude of 0.020 m.

(a) Calculate the speed of the wave, neglecting the mass of the wire.

(b) Write down an equation that describes the vertical displacement of the wave y as a function of horizontal distance x from the pin and time t . Assume a travelling wave with no reflections. Insert numerical values for any constants in the equation.

(c) What is the lowest frequency of the vibrating pin that will generate a standing wave in the wire?

(5 marks)

Solution (a) The speed is given by

Fvμ

=

where 10.100kg.mμ −= and ( )( )5.00 9.80 49.0 N.F m g= = = Therefore

149 22.1m.s .

0.10v −= =

(2 marks) (b)

( ) ( ), cosy x t A k x tω= ± For this situation we have 0.020 mA =

12 2 (40.0) 251.3 251rad.s .fω π π −= = = =

1251.3 11.4 rad.m .

22.1k

vω −= = =

And hence

( ) ( ), 0.020cos 11.4 251y x t x t= ± (student is free to use plus or minus in the cos argument as well as any additional phase/offset).

(½ mark for each of , ,A kω and also ½ mark for correct statement of ( , )y x t ) (c) The fundamental frequency is given by

( )( )

22.1 11.1Hz2 2 1.0vfL

= = = .

(1 mark)

ADV_Q06

Semester 1, Y2011 Question 6 (a) Explain what is meant by the term resonance and give an example of resonance in a real

mechanical system.

(b) What are two physical properties of your chosen system that determine its resonant frequency?

(c) What energy source drives the resonance in your chosen system?

(d) What physical process prevents the amplitude from becoming even larger than it is?

(5 marks)

Solution (a) Resonance is a phenomenon that refers to an oscillation that has a large amplitude because it is being driven at its natural frequency. Some examples are given below. Note that resonance can also occur with random excitation (white noise), such as wind on bridge, finger on wine glass, blowing over a bottle. That definition is ok, but the example should match the explanation for full marks. In other words, they should lose a mark for saying resonance occurs when an object is driven at its natural frequency and then gives an example that is randomly excited (wind on bridge, finger on wine glass, blowing over a bottle). (2 marks) (b) The two factors will generally relate to (i) the physical size of the system and (ii) its mass/density (see below). (2 marks) (c) The energy source will be some external driving agent (see below). Examples, with (answer to b)(and answer to c): (½ mark) (d) Damping limits the amplitude of the resonance. (½ mark) Examples

Example (a)

Physical Properties (b)

Energy Source (c)

Damping (d)

Child pushed on swing

• Length of swing • Strength of gravity not

mass of child Person pushing Friction (at pivot)

Bridge oscillating at large amplitude in wind or people walking

• Thickness/width of span

• material Wind or people Internal friction,

heating of material

Wine glass excited by loud speaker

• thickness of glass • diameter of glass loudspeaker Internal friction,

heating of material

ADV_Q07

Semester 1, Y2011 Question 7 You are designing a rocket for use in deep space, well away from any significant gravitational fields. Initially, the total mass of the rocket consists of the rocket itself, of mass M , plus 10 blocks of fuel, each with a mass m . To propel the rocket forward, the blocks of fuel are ejected out of the back at a speed v relative to the rocket.

On your first test flight, you start from rest and eject all of the fuel at once.

(a) Using the conservation of momentum, show that the velocity of the rocket after the fuel is ejected, 10V is given by:

1010 .mV vM

=

(b) Show that the impulse achieved through the ejection of all 10 blocks of fuel is the same whether they are ejected individually, or all at once.

Based on part (b), for your second run you start again from rest, loaded with 10 blocks of fuel of mass m , but now eject the fuel blocks individually at regular intervals in time. Again the fuel is ejected with a velocity v with respect to the rocket.

(c) Show that after the first block of fuel is ejected, the velocity of the rocket, v1, is given by:

1 .9

mv vM m

=+

(d) Remembering that the fuel speed, v , is relative to the rocket, show that the rocket's velocity after the second and then third blocks of fuel are ejected, 2v and 3v respectively, are given by:

v2 =m

M + 8m2v − v1( )

( )3 1 23 .7

mv v v vM m

= − −+

When all 10 blocks of fuel are ejected, the final velocity of the rocket is (you do not need to prove this):

v10 =mM

10v − v1 − v2 − v3 − v4 − v5 − v6 − v7 − v8 − v9( )

where iv is the velocity after the thi block has been ejected. Note that this is clearly different to the velocity you calculated in part (a).

(e) Although you showed in part (b) that the impulse does not depend upon the way the blocks were ejected, briefly explain, using physical principles, why the resultant velocity depends upon the method of ejection you adopt.

(10 marks)

Solution (a) Momentum is conserved so f ip p= . (1 mark) Initially, nothing is moving and so 0ip = . Finally,

10

10

10

10fp M V mv

mV vM

= −

⇒ =

(1 mark) (b)

J p F dtΔ = Δ = ∫ . (1 mark) All blocks have the same final velocity with respect to the rocket. The pΔ of each block is independent of the details of the force and the time, so long as the integral is the same. If ejected all together:

10 10p mvΔ = − . (1 mark) If ejected one at a time: The momentum of each block before ejection is rmv+ where rv is the velocity of the rocket. The momentum after ejection is: ( )rm v v− . Hence the change in momentum of each block is: ( )i r rp m v v mv mvΔ = − − = − which is independent of rv . So:

10tot ip p mvΔ = Δ = −∑ Which is the same as 10pΔ . (1 mark) (c) Ejection of the first block

1

1

0 ( 9 )

9

i f

i f

p and p M m v

mp p v vM m

= = +

= ⇒ =+

(1 mark) (d) Ejection of the second block Momentum is still conserved so

2 1

2 1

2 1

( 8 ) ( ) 0 (rocket) (1st block) (2nd block)

( 8 ) (2 )

(2 )( 8 )

M m v mv m v v

M m v m v vmv v v

M m

+ − − − =

⇒ + = −

⇒ = −+

(1 mark) Ejection of the third block Momentum is still conserved so

2 1 2

2 1 2

2 1 2

( 7 ) ( ) ( ) 0 (rocket) (1st block) (2nd block) (3rd block)

( 7 ) (3 )

(3 )( 7 )

M m v mv m v v m v v

M m v m v v vmv v v v

M m

+ − − − − − =

⇒ + = − −

⇒ = − −+

(1 mark) (e)

v10 =mM

10v − v1 − v2 − v3 − v4 − v5 − v6 − v7 − v8 − v9( )

Even though the impulse is the same for each ejection of a block some is used to change the momentum of the remaining fuel when the blocks are ejected individually. When the ten blocks are ejected all at once, all of the momentum is transferred to the rocket. (2 marks)

ADV_Q08

Semester 1, Y2011 Question 8

A solid uniform ball of radius R and mass m (moment of inertia 225

I m R= ) rolls without

slipping up a hill. It has an initial velocity of 125.0m.s− at the bottom of the hill which is 28.0 m high as shown in the diagram.

At the top of the hill, it is moving horizontally and then goes over a vertical cliff.

(a) Show that the initial kinetic energy totalK of the ball at the bottom of the hill is given by

2710total BK mv=

where Bv is the speed of the ball at the bottom of the hill.

(b) Determine the speed of the ball at the top of the hill.

(c) Show that the rotational kinetic energy of the ball at the top of the hill is given by

215rotation TK mv=

where Tv is the speed of the ball at the top of the hill.

(d) Determine the speed of the ball just before it hits the ground. You can neglect any effect of air resistance.

(e) This speed is greater than the initial speed when it set off up the hill. Does this mean that the ball has somehow gained energy? Explain your answer.

(10 marks)

Solution (a) The initial kinetic energy is:

2 2

22 2

2

1 12 21 1 22 2 57

10

total translation rotation

B

BB

B

K K K

mv I

vmv m RR

mv

ω

= +

= +

⎛ ⎞⎛ ⎞= + ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

=

where we have used Bv Rω= for rolling without slipping. (2 marks) (b) At the top of the hill, the kinetic energy has been reduced by an amount

gU m g hΔ = . (1 mark) Its speed Tv at the top of the hill is then determined by:

2 27 7

10 10T Bmv mv m g h= −

Hence

2 2 10

7T Bv v g h= −

which evaluates to

( ) ( )( )

( )

22

1

1025.0 9.8 28.0 2337

15.26 15.3 m.s .

T

T

v

v −

= − =

⇒ =

The speed at the top of the hill is 15.3 m/s. (2 marks) (c) The rotational kinetic energy at the top of the hill is

22 2

2

1 1 22 2 5

15

Trotation

T

vK I mRR

mv

ω ⎛ ⎞⎛ ⎞⎛ ⎞= = ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

=

and this stays constant while the ball is in free-fall. (1 mark) (d) All of the work done by gravity as the ball falls back to the ground results in an increase in its translational kinetic energy.

( )

2 2

22 2 2

1

1 12 2

42 25.0 156.87

28.0m.s .

f T

f T B

f

m v mv m g h

v v g h v g h

v −

= +

= + = + = +

⇒ =

The speed just before it hits the ground is 28.0 m/s. (2 marks) (e) The ball has not gained energy; it has the same energy as at the beginning. The rotational kinetic energy is the same as when it was at the top of the hill, which is less than the rotational kinetic energy at the start. Given that the total kinetic energy is the same as at the start, it has more translational kinetic energy than at the start, and therefore a greater speed. (2 marks)

REG_Q09=ADV_Q09

Semester 1, Y2011 Question 9

A sample of 2.00 mol of Helium gas (assumed to be ideal, monatomic gas) has a volume of 30.0326m , a pressure of 1.50 atm , and a temperature of 25.0 C (State A). A physicist firstly

heats the gas at constant volume, adding 41.50 10 J× of heat, to reach State B. He then continues heating and allows the gas to expand at constant pressure to twice its original volume, reaching State C. Answer the following questions, in each case making it clear how you obtained your results in terms of known parameters such as , , etcA A Ap V T .

(a) Sketch a pV diagram for the whole process showing the positions of States A, B and C. For now, you need not calculate any unknown values for p and V .

(b) Write down an expression for the molar heat capacity of the gas at constant volume and use it to calculate the temperature of the gas at State B.

(c) Calculate the pressure and the temperature of the gas at State C.

(d) Calculate the amount of work done by the gas.

(e) Calculate the change in internal energy of the gas for the whole process from A to C.

(10 marks)

Solution Pressure at State A is ( )( )5 51.50 1.013 10 1.52 10 Pa.× = × Temperature at State A is 273.15 25.0 298.2 K+ =

The table below shows the values (or relationships) given in the problem and those calculated later in the solution (highlighted text). Pressure Volume Temperature State A 51.520 10 Pa× 30.0326 m 298.2 K

State B 54.59 10 Pa× Same as A

30.0326 m 899.6 K

State C Same as B

54.59 10 Pa× Double B

30.0652 m 1800 K

(a) The processes are shown in the following pV diagram. The actual values for p and V are not required.

(1 mark) (b) Molar heat capacity for an ideal monatomic gas at constant volume is given by

3 .2VC R=

(1 mark)

( )( )( )41.50 10 601.4 K

2.00 1.5 8.314VV

dQdQ nC dT dTnC

×= ⇒ = = =

So temperature at State B is ( )601.4 273.15 25.0 899.6K+ + = (2 marks) (c) Calculate the pressure at State B Bp from the ideal gas law as follows:

( )( )( ) ( )5 52.00 8.314 899.6

4.589 10 4.59 10 Pa.0.0326

BB B B B

B

n RTp V n RT pV

= ⇒ =

= = × ×

The change from State B to State C is at constant pressure and so

54.59 10 Pa.C Bp p= = ×

(1 mark) Calculate the temperature at State C from the ideal gas law as follows:

( )( )( )( )( ) ( )

54.589 10 2 0.03261799 1800 K.

2.00 8.314

C CC C C C

p Vp V n RT Tn R

= ⇒ =

×= =

(1 mark) (d) Work done by the system is given by W p dV= ∫ We can consider the two processes separately as follows: State A →State B This is isochoric (constant volume) and there is no work done by the gas as 0.dV = (1 mark) State B → State B This is isobaric (constant pressure) and the work done bt the gas can be written

( )

( )( ) ( )5 44.589 10 0.0326 1.496 1.50 10 J.

C C

B B

V V

B B C B B AV V

W p dV p dV p V V p V= = = − =

= × = ×

∫ ∫

(1 mark) (e) Internal energy of an ideal gas is determined only by the change in temperature.

( ) ( )( )( )( )4

3 1.5 2.00 8.314 1799 298.22

3.74 10 J.

V C AdU nC dT n R T T= = − = −

= ×

(2 marks)

ADV_Q10

Semester 1, Y2011 Question 10 (a) A Volkswagen Passat has a six-cylinder Otto-cycle engine with compression ratio

10.6r = . The pV-diagram for the Otto cycle is:

The initial pressure and temperature of the air-fuel mixture at point (a) are:

48.50 10 Paap = × and 300KaT = . The heat added to each cylinder in each cycle is 200 J . Assuming the initial volume occupied by the fuel mixture in each cylinder is

4 35.00 10 maV −= × , and that the fuel mixture behaves like an ideal gas, with 1 120.5 J.mol .KVC − −= and 1.40γ = , calculate the following:

(i) The number of moles n of the fuel mixture.

(ii) The maximum temperature reached when the fuel is ignited in step 2.

(iii) The thermal efficiency of each cycle, using 111e

rγ −= − .

(iv) The efficiency of a Carnot-cycle engine operating between the same maximum and minimum temperatures.

(b) A sample of 2.0 moles of an ideal gas occupies a volume V . The gas expands isothermally and reversibly to a volume 3 .V

(i) Does the velocity distribution of the molecules change? Briefly explain your answer.

(ii) Calculate the change in entropy using the macroscopic approach.

(iii) Calculate the change in entropy using the microscopic approach.

(10 marks)

Solution

(a)(i)

( )( )( )( )

4 42

8.50 10 5.00 101.70 10 mol

8.314 300pVnRT

−−

× ×= = = ×

(1 mark) (a)(ii)

( )H V c bQ nC T T= − and 1b aT T rγ −=

( )( )( )

( )( )0.4

2

3

1

200300 10.6

1.70 10 20.5

469.5 771.3K=1.24 10 K

Hc a

V

QT T rnC

γ

−

−⇒ = +

= +×

= +

×

(1 mark) (a)(iii)

( )0.41 11 1 1 0.3889

2.57110.60.611

ottoe = − = − = −

=

(1 mark) (a)(iv)

3001 1 0.7581240

Ccarnot

H

TeT

= − = − =

(2 marks) (b)(i) No, because the velocities depend only on temperature, which remains unchanged. (1 mark) (b)(ii)

QST

Δ = and 2

1

ln VQ dQ dW pdV nRTV

⎛ ⎞= = = = ⎜ ⎟

⎝ ⎠∫ ∫ ∫ .

Therefore ( ) ( )( )( ) 1ln 3 2.0 8.314 1.099 18J.K .S n R −Δ = = =

(2 marks) (b)(iii)

( )2

1

ln ln 3N

A

w RS kw N

⎛ ⎞Δ = =⎜ ⎟

⎝ ⎠

since w is the number of microstates occupied by N molecules, where AN n N= and n is the number of moles of gas. Therefore

( ) 1ln 3 18J.KS nR −Δ = = which is the same result as for the macroscopic method. (2 marks)

ADV_Q11

Semester 1, Y2011 Question 11 The following differential equation describes an oscillator for which the damping force is proportional to velocity.

2

2 .d x dxm k x bdt dt

= − −

(a) Briefly explain the physical meanings of x , m , b and k in the above equation. We are told that a solution of this equation is

( ) ( )cos .Btx t Ae tω−=

(b) Briefly explain the physical meanings of A , B and ω in the above equation.

(c) Show that

.2bBm

=

Hint: consider the complex quantity ( ) B t i tz t Ae e ω−= whose real part is ( )x t .

(d) Would you expect this system to exhibit chaotic behaviour? Give reasons for your answer.

(10 marks)

Solution (a) x and m are the position and mass of the oscillating object, b measures the strength of the damping force and k measures the strength of the restoring force. (2 marks) (b) A is the amplitude of the oscillation, B is the reciprocal of the damping time, and ω is the angular frequency of the oscillation (which, by the way, is less than the frequency of the undamped oscillator).

(½ mark for each of A and ω; 1 mark for B) (c) We need to solve the equation

2

2 0,d z dzm b kzdt dt

+ + =

where z is complex. If z is a solution of equation 1 then ( )zℜ is a solution of the oscillator equation. This follows because the differential equation is linear ( z and its derivatives only appear in the zeroth or first power). We are told that the solution of Equation 1 is

( ) .B i tBt i tz Ae e Ae ωω − +−= = The derivatives of z are

( )dz B i zdt

ω= − +

(1 mark) and

( )

( )

22

2 2 2 .

d z B i zdt

B B i z

ω

ω ω

= − +

= − −

(1 mark)

Putting these into Equation 1 gives ( ) ( )2 2 2 0m B B i z b B i k zω ω ω− − + − + + =

(1 mark) This will be true when both the real and the imaginary parts of the left hand side are zero. The latter gives us 2 0m B bω ω− + = And hence

,2bBm

=

as desired.

(1 mark) (d) No, it would not display chaos. It has only two degrees of freedom, and there is no non-linearity in the differential equation. Note that loss of energy (through damping) does not rule out chaotic behaviour. (2 marks – deduct a mark if damping is given as a reason)

REG_Q12=FND_Q12=ADV_Q12

Semester 1, Y2011 Question 12 Richard wishes to test the Doppler effect for sound waves. To do so, he attaches a sound wave generator to the arm of a large centrifuge (normally used to train astronauts and pilots to withstand high “g” forces) as shown in the diagram below. The distance between the centre of rotation and the sound generator is 10.0 m . The sound generator produces a single tone of frequency 2000 Hz and the centrifuge is set to make a complete revolution once every 2.0s . The speed of sound in air is 1344 m.s− . A sound detector is placed at rest at a large horizontal distance (much larger than the size of the centrifuge) to the right of the centrifuge and is used to measure the frequency and wavelength of the sound produced by the generator.

(a) Calculate the tangential speed of the sound generator as the centrifuge rotates.

(b) Calculate the wavelength of the sound measured by the detector before the centrifuge begins to rotate.

(c) The detector finds that the frequency of the sound wave is changing as the centrifuge rotates. Explain why this occurs.

(d) Describe how the detected sound frequency changes as the centrifuge rotates, indicating at which point(s) (numbered 1, 2, 3, 4 in the above diagram) the frequency is highest and at which point(s) it is lowest.

(e) What is the value of the highest frequency measured by the detector?

(f) What is the wavelength of the sound measured by the detector when the frequency is highest?

(10 marks)

Solution (a) Circumference of the circular path of the sound generator is:

( )2 2 10.0 62.83m.C Rπ π= = = This is covered in one rotation period of 2.0s.T = Hence the tangential speed of the sound generator is:

( )1 12 31.42m.s 2 sig figs 31m.stRv

Tπ − −= = .

(1 mark) (b) Frequency of the sound at rest is 2000 Hz. Wavelength is calculated from

( )344 m 0.172m 2 sig figs 0.17m .2000

vv ff

λ λ= ⇒ = = =

(1 mark) (c) The velocity of the generator relative to the detector is constantly changing, so that the frequency of the sound is Doppler-shifted, and this shift is constantly changing as the generator rotates. (2 marks) (d) The frequency observed is highest when the generator is approaching the detector (position 3) and lowest when the generator is moving directly away from the detector (position 1). (1 mark for explanations, 1 mark for positions) (e) The highest frequency is observed when the generator is moving towards the detector at position 3. At this time the velocity of the generator towards the detector is that given in part (a).

LL S

S

v vf fv v+

=−

where v is the velocity of sound, Lv is the velocity of the listener (the detector) relative to the medium (air), and Sv is the velocity of the source relative to the medium (air), and sf is the frequency of the source. For this situation we have

10 m.sLv −= 131.42 m.sSv −= +

1344m.sv −= Therefore

( ) ( )344 02000 Hz 2200 Hz.

344 31.42Lf+

= =−

(2 marks)

(f) Wavelength is given by

( ) ( )344 31.420.156 m 2 sig figs 0.16m .

2000S

S

v vf

λ−−

= = = .

Alternatively 344 m 0.16 m.2200L

vf

λ = = =

(2 marks)